Chapter 3: Dielectric And Magnetic Materials

Example 3.17.1,Page number 3-35

In [1]:
import math

#given data

A=650*10**-6                             #area
d=4*10**-3                               #seperation of plate
Q=2*10**-10                              #charge
er=3.5                                   #relative permitivity

e0=8.85*10**-12                          #absolute permitivity

V=(Q*d)/(e0*er*A)

print"voltage across capacitor =",round(V,4),"Volt"

voltage across capacitor = 39.7343 Volt

Example 3.17.2,Page number 3-36

In [4]:
import math

#given data

A=2000*10**-6                           #area
d=0.5*10**-6                            #seperation of plate
er=8.0                                    #relative permitivity
e0=8.85*10**-12                         #absolute permitivity

C=(e0*er*A)/d

print"capacitance for capacitor =","{0:.3e}".format(C),"Faraday"

capacitance for capacitor = 2.832e-07 Faraday

Example 3.17.3,Page number 3-36

In [6]:
import math

#given data
E=1000                                  #electric field
P=4.3*10**-8                            #polarization
e0=8.854*10**-12                        #absolute permitivity
er=(P/(e0*E))+1                         #as P/E=e0(er-1)

print"relative permittivity =",round(er,4)

relative permittivity = 5.8566

Example 3.17.4,Page number 3-36

In [8]:
import math

#given data

#As C=e0*er*A/d

e0=math.e                               #absolute permitivity

Ag=1l

Ap=Ag                                   #Assuming Area of glass plate and plastic film is same

#for glass

erg=6                                   #relative permitivity

dg=0.25                                 #thickness

Cg=e0*erg*Ag/dg

#for plastic film

erp=3                                   #relative permitivity

dp=0.1                                  #thickness

Cp=e0*erp*Ap/dp

m=Cg/Cp

print"since Cg/Cp=",m

print"plastic film holds more charge"

since Cg/Cp= 0.8
plastic film holds more charge

Example 3.17.5,Page number 3-37

In [13]:
import math

#given data

N=2.7*10**25                             #no of atoms per m**3
er=1.0000684                            #dielectric constant of He atom at NTP
e0=8.854*10**-12                         #absolute permitivity

a=e0*(er-1.0)/N                           #electronic polarizability

print"1) electronic polarizability=","{0:.3e}".format(a)

R=(a/(4*3.1472*e0))**(1.0/3)                  #radius of helium atom

print"2) radius of He atoms =","{0:.3e}".format(R),"meter"

1) electronic polarizability= 2.243e-41
2) radius of He atoms = 5.860e-11 meter

Example 3.17.6,Page number 3-37

In [14]:
import math

er=1.000014                             #dielectric constant of He atom at NTP
Xe=er-1.0                                 #electric susceptibility

print"electric susceptibility =",(Xe)

electric susceptibility = 1.4e-05

Example 3.17.7,Page number 3-37

In [16]:
import math

#given data

T=300                                   #temperature of paramagnetic material
X=3.7*10**-3                             #susceptibility of material

C=X*T                                   #using Curie's law

T1=250                                  #temperature
T2=600                                  #temperature

u1=C/T1                                 #relative permeability of material at 250k

u2=C/T2                                 #relative permeability of material at 350k

print"relative permeability at temp 250K=","{0:.3e}".format(u1)

print"relative permeability at temp 600K =","{0:.3e}".format(u2)

relative permeability at temp 250K= 4.440e-03
relative permeability at temp 600K = 1.850e-03

Example 3.17.8,Page number 3-38

In [18]:
import math

#given data

u=0.8*10**-23                            #magnetic dipole moment of an atom 
B=0.8                                   #magnetic field
K=1.38*10**-23                           #boltzmann constant

T=(2*u*B)/(3*K)                         #temperature

print"Temperature at which average thermal energy of an atom is equal to magntic energy=",round(T,4),"K"

Temperature at which average thermal energy of an atom is equal to magntic energy= 0.3092 K

Example 3.17.9,Page number 3-38

In [20]:
import math

#given data

B=0.5                                   #magnetic field
t=27                                    #temperature in degree celcius
T=273+t                                 #temperature in kelvin

u0=4*math.pi*10**-7                          #permeability of free space

C=2*10**-3                               #Curie's constant

M=(C*B)/(u0*T)                          #magnetization of material

print"magnetization of paramagnetic material =",round(M,4),"A/m"

magnetization of paramagnetic material = 2.6526 A/m

Example 3.17.10,Page number 3-38

In [21]:
import math

#given data

u0=4*math.pi*10**-7                          #permeability of free space
B=10.9*10**-5                            #flux density

H=B/u0                                  #magnetic field

print"Horizontal component of magnetic field =",round(H,4),"A-m"

Horizontal component of magnetic field = 86.7394 A-m

Example 3.17.11,Page number 3-39

In [22]:
import math

#given data

phi=5.9*10**-3                           #magnetic flux
ur=900                                  #relative permeability of material
n=700                                   #number of turns

u0=4*math.pi*10**-7                     #permeability of free space

A=60*10**-4                             #cross section area of ring

l=2                                     #mean circumference of ring

B=phi/A                                 #flux density

H=B/(u0*ur)                             #magnetic field

At=H*l                                  #Amp-turns required

I=At/n                                  #current required

print"Current required to produce a flux=",round(I,4),"Amp"

Current required to produce a flux= 2.4842 Amp

Example 3.17.12,Page number 3-39

In [23]:
import math

#given data

phi=2.7*10**-3                           #magnetic flux
A=25*10**-4                              #cross section area of ring
r=25*10**-2                              #mean circumference of ring
la=10**-3                                #air gap

ur=900                                  #relative permeability of material
n=400                                   #number of turns

u0=4*math.pi*10**-7                          #permeability of free space

d=40*10**-2                              #mean diameter of ring

li=2*math.pi*r                              #mean circumference of ring

B=phi/A                                 #flux density

#for air gap

Ha=B/(u0)                               #magnetic field for air gap

#for iron ring

Hi=B/(u0*ur)                            #magnetic field for iron ring

#therefore, Amp turn in air gap

Ata=Ha*la                               #Amp-turns required

#therefore, Amp-turn in ring

Ati=Hi*li                               #Amp-turns required

#therrfore total mmf required

mmf=Ata+Ati

#Current required

I=mmf/n                                  #current required

print"Current required =",round(I,4),"Amp"

Current required = 5.8986 Amp

Example 3.17.13,Page number 3-40

In [3]:
import math

#given data

n1=10                                   #no of turns per cm
i=2                                     #current
B=1                                     #flux density

u0=4*math.pi*10**-7                     #permeability of free space

n=n1*100                                #no turns per m

H=n*i

print"1) magnetic intensity =",round(H,4),"Amp-turn/meter"

#calculation for magnetization

I=B/u0-H

print"2) magnetization =","{0:.3e}".format(I),"Amp-turn/meter"

#relative permeability

ur=B/(u0*H)

print"3) Relative Permeability of the ring =",(int(ur))

1) magnetic intensity = 2000.0 Amp-turn/meter
2) magnetization = 7.938e+05 Amp-turn/meter
3) Relative Permeability of the ring = 397

Example 3.17.14,Page number 3-40

In [4]:
import math

#given data

m=40                                    #wt of the core
d=7.5*10**3                              #density of iron
n=100                                   #frequency

V=m/d                                   #volume of the iron core

E1=3800*10**-1                           #loss of energy in core per cycles/cc

E2=E1*V                                 #loss of energy in core per cycles

N=60*n                                  #no of cycles per minute

E=E2*N                                  #loss of energy per minute

print"Loss of energy per minute =",(E),"Joule"

Loss of energy per minute = 12160.0 Joule

Example 3.17.15,Page number 3-40

In [9]:
import math

#given data

l=30*10**-2                             #length of ring
A=1*10**-4                              #cross section area of ring
i=0.032                                 #current

phi=2*10**-6                             #magnetic flux

u0=4*math.pi*10**-7                      #permeability of free space

N=300                                    #no of turns in the coil

#1) flux density

B=phi/A                                 #flux density

print"1) Flux density in the ring =",(B),"Wb/m**2"

#2) magnetic intensity of ring

n=N/l                                   #no of turns per unit length

H=n*i                                   #magnetic intensity

print"2) magnetic intensity =",(H),"Amp-turn/meter"

#3) permeability and relative permeability of the ring

u=B/H

print"3) Permeability of the ring =","{0:.3e}".format(u),"Wb/A-m"

ur=u/u0

print"4) Relative Permeability of the ring =",round(ur,4)

#4)Susceptibility

Xm=ur-1

print"5) magnetic Susceptibility of the ring =",round(Xm,4)

1) Flux density in the ring = 0.02 Wb/m**2
2) magnetic intensity = 32.0 Amp-turn/meter
3) Permeability of the ring = 6.250e-04 Wb/A-m
4) Relative Permeability of the ring = 497.3592
5) magnetic Susceptibility of the ring = 496.3592

Example 3.17.16,Page number 3-41

In [10]:
import math

#given data

E=3000                                  #loss of energy per cycle per cm**3
m=12*10**3                               #wt of the core
d=7.5                                   #density of iron
n=50                                    #frequency

V=m/d                                   #volume of the core

El=E*V*n*60*60                          #loss of energy per hour

print"Loss of energy per hour =",(El),"Erg"

Loss of energy per hour = 8.64e+11 Erg

Example 3.17.17,Page number 3-41

In [11]:
import math

#given data

n=50                                    #frequency
V=10**-3                                #volume of the specimen

#Area of B-H loop

A=0.5*10**3*1

P=n*V*A

print"Hysteresis power loss =",(P),"Watt"

Hysteresis power loss = 25.0 Watt

Example 3.17.18,Page number 3-42

In [13]:
import math

#given data

phi=1.5*10**-4                           #magnetic flux

ur=900                                  #relative permeability of material

n=600                                   #number of turns

u0=4*math.pi*10**-7                     #permeability of free space

A=5.8*10**-4                             #cross section area of ring

d=40*10**-2                              #mean diameter of ring

li=math.pi*d                             #mean circumference of ring

la=5*10**-3                              #air gap

B=phi/A                                  #flux density

#for air gap

Ha=B/(u0)                               #magnetic field for air gap

#for iron ring

Hi=B/(u0*ur)                            #magnetic field for iron ring

#therefore, Amp turn in air gap

Ata=Ha*la                               #Amp-turns required

#therefore, Amp-turn in ring

Ati=Hi*li                               #Amp-turns required

#therrfore total mmf required

mmf=Ata+Ati

#Current required

I=mmf/n                                  #current required

print"Current required =",round(I,4),"Amp"

Current required = 2.194 Amp

Example 3.17.19,Page number 3-42

In [15]:
import math

#given data

la=1*10**-2                              #air gap
r=0.5                                    #radius of ring
A=5*10**-4                               #cross section area of ring
i=5                                      #current
u=6*10**-3                               #permeability of iron
u0=4*math.pi*10**-7                      #permeability of free space
N=900                                    #no of turns in the coil

#let reluctance of iron ring with air gap be S

S=la/(u0*A)+(2*math.pi*r-la)/(u*A)

print"1) Reluctance =","{0:.3e}".format(S),"A-T/Wb"

mmf=N*i

print"2) m.m.f =",(mmf),"Amp-turn"

1) Reluctance = 1.696e+07 A-T/Wb
2) m.m.f = 4500 Amp-turn

Example 3.17.20,Page number 3-43

In [16]:
import math

#given data

#the magnetization force is given by,
#H=NI/l

H=5*10**3                                #coercivity of bar magnet
l=10*10**-2                              #length of solenoid
N=50                                     #number of turns

I=l*H/N

print"current =",(I),"Ampere"

current = 10.0 Ampere

Example 3.17.21,Page number 3-43

In [20]:
import math

#given data

ur=380                                  #relative permeability of air
u0=4*math.pi*10**-7                     #permeability of free space
A=5*10**-4                              #cross section area of ring
n=200                                   #number of turns
d=20*10**-2                             #mean diameter of ring

l=math.pi*d                             #mean circumference of ring

phi=2*10**-3                            #magnetic flux

S=l/(u0*ur*A)                           #reluctance

#using ohm's law for magnetic circuit

#phi=N*I/S

I=S*phi/n

print"1) Reluctance =","{0:.3e}".format(S),"A-T/Wb"
print"2) current =",round(I,4),"Ampere"

1) Reluctance = 2.632e+06 A-T/Wb
2) current = 26.3158 Ampere

Example 3.17.22,Page number 3-43

In [22]:
import math

#given data

ur=1                                    #relative permeability of air
u0=4*math.pi*10**-7                     #permeability of free space
A=6*10**-4                              #cross section area of torroid
n=500                                   #number of turns
r=15*10**-2                             #radius of torroid
I=4                                     #current in coil
l=2*math.pi*r                           #mean circumference of torroid
MMF=n*I

print"1) MMF (NI) =",(MMF),"AT"

R=l/(u0*ur*A)                           #Reluctance

print"2) Reluctance (R) =","{0:.3e}".format(R),"AT/Wb"

phi=MMF/R                               #flux

print"3) Magnetic flux =",(phi),"Wb"

B=phi/A                                 #flux density

print"4) Flux density =","{0:.3e}".format(B),"Wb/m**2"

H=B/(u0*ur)                             #magnetic field intensity

print"5) Magnetic field intensity =",round(H,4),"A/m"

1) MMF (NI) = 2000 AT
2) Reluctance (R) = 1.250e+09 AT/Wb
3) Magnetic flux = 1.6e-06 Wb
4) Flux density = 2.667e-03 Wb/m**2
5) Magnetic field intensity = 2122.0659 A/m

Example 3.17.23,Page number 3-44

In [24]:
import math

#given data

phi=10**-3                              #magnetic flux
ur=1000                                 #relative permeability of iron
u0=4*math.pi*10**-7                     #permeability of free space
A=5*10**-4                              #cross section area of ring
la=2*10**-3                             #air gap
d=20*10**-3                             #mean diameter of ring

li=math.pi*d-la                         #mean circumference of ring

#using KVL for magnetic circuit

#AT(total)=AT(iron)+AT(air gap)

ATt=(phi/(u0*A))*((li/ur)+la)

print"Number of Ampere-Turns required =",round(ATt,0)

Number of Ampere-Turns required = 3280.0

Example 3.17.24,Page number 3-44

In [26]:
import math

#given data

X=0.5*10**-5                            #susceptibility of material

H=10**6                                 #magnetic field strength

I=X*H                                   #intensity of magnetization

u0=4*math.pi*10**-7                     #permeability of free space

B=u0*(H+I)                              #flux density

print"1) intensity magnetization =",(I),"Amp/m"

print"2) flux density in the material =",round(B,4),"wb/m**2"

1) intensity magnetization = 5.0 Amp/m
2) flux density in the material = 1.2566 wb/m**2
In [ ]: